# Thread: Geometric/arithmetic progressions in geometric trig

1. ## Geometric/arithmetic progressions in geometric trig

In triangle ABC, AB = BC, and BD is an altitude. Point E is on the extension of AC such that BE = 10. The values of tan<CBE, tan<DBE, and tan<ABE form a geometric progression, and the values of cot< DBE, cot< CBE, cot< DBC form an arithmetic progression. What is the area of triangle ABC?

2. ## Re: Geometric/arithmetic progressions in geometric trig

Well, so far I got, where X is CBE, and K is a constant in the geometric progression, and N is the constant in the arithmetic progression-

1/XK = 1/X + N = 1/[X(K-1)]+2N

Any ideas where to go from this?

3. ## Re: Geometric/arithmetic progressions in geometric trig

My intuition is that you want the triangle such that AD = DB = BE = 4, BD = 6

Then the area is trivially (4 * 6)/2 = 12

This needs to be shown, and could be wrong, it's just intuition.

The reason I think this is because the length of the line the angles subtend at AE become 4, 8, 12. But it's just a hunch.

To solve it you've got to come up with expressions for the tangents of those angles using the geometry and similar for the cotangents. BD and AC are the things you can vary to ensure the conditions mentioned.

you're going to be the similar triangle master by the end of the day

just to review

geometric progression is: a, ar, ar2, ar3 ...

arithmetic progression is: a, a+b, a+2b, a+3b ...